急证:sin t + cos t - 2/[1+tan^2 (t/2) ]

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急证:sint+cost-2/[1+tan^2(t/2)]急证:sint+cost-2/[1+tan^2(t/2)]急证:sint+cost-2/[1+tan^2(t/2)]sint=2tan(t/2

急证:sin t + cos t - 2/[1+tan^2 (t/2) ]
急证:sin t + cos t - 2/[1+tan^2 (t/2) ]

急证:sin t + cos t - 2/[1+tan^2 (t/2) ]
sin t = 2tan (t/2) /[1+tan^2 (t/2) ]
cos t = [1-tan^2 (t/2) ] /[1+tan^2 (t/2) ]
以上为万能公式 ,代入知
sin t + cos t - 2/[1+tan^2 (t/2) ]

2/[1+tan^2 (t/2) ]=2cos^2 (t/2)=cos t+1
所以sin t + cos t - 2/[1+tan^2 (t/2) ]=sin t + cos t - (cos t+1)=sin t - 1<=0

2/[1+tan^2 (t/2) ]=2cos^2 (t/2)=cos t+1
所以sin t + cos t - 2/[1+tan^2 (t/2) ]=sin t + cos t - (cos t+1)=sin t - 1<=

使用万能公式就可以一步证明了:
sin t = 2tan (t/2) /[1+tan^2 (t/2) ]
cos t = [1-tan^2 (t/2) ] /[1+tan^2 (t/2) ]
以上为万能公式 ,代入知
sin t + cos t - 2/[1+tan^2 (t/2) ] <=0

原式=sin t cos t–2cos∧2(t/2)=2sin(t/2)cos(t/2) cos∧2(t/2)-sin∧2(t/2)-2cos∧2(t/2)=-【sin∧2(t/2)-2sin(t/2)cos(t/2) cos∧2(t/2)】=-【sin(t/2)-cos(t/2)】∧2≤0 应该看的懂吧,最后的形式很直接就可以看出其范围了,t取任意值都满足。

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